【POJ】1330 Nearest Common Ancestors ——最近公共祖先（LCA）

Nearest Common Ancestors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18136		Accepted: 9608

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:

In the figure, each node is labeled with an integer from {1,
2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node
y if node x is in the path between the root and node y. For example,
node 4 is an ancestor of node 16. Node 10 is also an ancestor of node
16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of
node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6,
and 7 are the ancestors of node 7. A node x is called a common ancestor
of two different nodes y and z if node x is an ancestor of node y and an
ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of
nodes 16 and 7. A node x is called the nearest common ancestor of nodes y
and z if x is a common ancestor of y and z and nearest to y and z among
their common ancestors. Hence, the nearest common ancestor of nodes 16
and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.

For other examples, the nearest common ancestor of nodes 2 and 3 is
node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and
the nearest common ancestor of nodes 4 and 12 is node 4. In the last
example, if y is an ancestor of z, then the nearest common ancestor of y
and z is y.

Write a program that finds the nearest common ancestor of two distinct nodes in a tree.

Input

The
input consists of T test cases. The number of test cases (T) is given in
the first line of the input file. Each test case starts with a line
containing an integer N , the number of nodes in a tree,
2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N.
Each of the next N -1 lines contains a pair of integers that represent
an edge --the first integer is the parent node of the second integer.
Note that a tree with N nodes has exactly N - 1 edges. The last line of
each test case contains two distinct integers whose nearest common
ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5

Sample Output

4
3

Source

Taejon 2002

 

题意：构建一棵树并给你一个询问，求这两个点的最近公共祖先。

题解：此题数据很水，暴力可过，tarjan最佳。

代码如下：

 #include <cstdio>
 #include <cstring>
 #include <vector>
 using namespace std;

 const int LEN = ;

 vector<int> vec[LEN];
 int uset[LEN];
 bool vis[LEN];
 bool root[LEN];

 void init(int n)
 {
     for(int i = ; i <= n; i++)
         vec[i].clear();
 }

 void makeset(int n)
 {
     uset[n] = n;
 }

 int findset(int x)
 {
     return x == uset[x] ? x : uset[x] = findset(uset[x]);
 }

 void unionset(int x, int y) //并查集操作
 {
     x = findset(x);
     y = findset(y);
     if (x == y)
         return;
     uset[y] = x;
 }

 void LCA(int u, int q1, int q2)
 {
     int v;
     makeset(u);
     for(int i = ; i < vec[u].size(); i++){
         v = vec[u][i];
         LCA(v, q1, q2);
         unionset(u, v); //后续遍历并合并集合
     }
     vis[u] = true;
     if (u == q1 && vis[q2] == true){ //如果访问到询问点，判断另外一个点是否被访问过，如果访问过则该点为最近公共祖先
         printf("%d\n", findset(q2));
         return;
     }
     else if (u == q2 && vis[q1] == true){
         printf("%d\n", findset(q1));
         return;
     }

 }

 int main()
 {
     int T, n, a, b, q1, q2;
     scanf("%d", &T);
     while(T--){
         memset(uset, , sizeof(uset));
         memset(vis, , sizeof(vis));
         memset(root, , sizeof(root));
         scanf("%d", &n);
         init(n);
         for(int i = ; i < n - ; i++){
             scanf("%d %d", &a, &b);
             vec[a].push_back(b);
             root[b] = true; //标注非根节点
         }
         scanf("%d %d", &q1, &q2);
         for(int i = ; i <= n; i++)
             if (root[i] != true){ //从根节点开始遍历
                 LCA(i, q1, q2);
                 break;
             }
        }
     return ;
 }